Q: The motion of a particle varies with time according to the relation y = a\sin \omega t + a\cos \omega t . Then

y=asin\omega t+acos\omega t\\ \quad =\frac { \sqrt { 2 } a }{ \sqrt { 2 } } (sin\omega t+cos\omega t)\\ \quad =\sqrt { 2 } a(\frac { 1 }{ \sqrt { 2 } } sin\omega t+\frac { 1 }{ \sqrt { 2 } } cos\omega t)\\ \quad =\sqrt { 2 } a(sin\omega t\quad cos\pi /4\quad +cos\omega t\quad sin\pi /4)\\ \quad =\sqrt { 2 } a\quad sin(\omega t+\frac { \pi }{ 4 } )\\ This\quad equation\quad satisfy\quad the\quad SHM\quad equation,\quad \frac { { d }^{ 2 }y }{ { d }^{ 2 }t } =-y\quad with\quad amplitude\quad \sqrt { 2 } a\quad .\quad \\

Q: A mass m= 100\ gms is attached at the end of a light spring which oscillates on a friction less horizontal table with an amplitude equal to 0.16\ meter and the time period equal to 2\ sec . Initially the mass is released from rest at t=0 and displacement x= -0.16\ meter . The expression for the displacement of the mass at any time (t) is

Given: m=100 gms\\a=0.16m\\T=2\ sec Standard equation for given condition x=a\cos \dfrac{2\pi}{T}t\Rightarrow x=-0.16\cos ( \pi t)

Question 1

The motion of a particle varies with time according to the relation  y = a\sin  \omega   t + a\cos  \omega t   . Then

Accepted Answer

y=asin\omega t+acos\omega t\ \quad =\frac { \sqrt { 2 } a }{ \sqrt { 2 }  } (sin\omega t+cos\omega t)\ \quad =\sqrt { 2 } a(\frac { 1 }{ \sqrt { 2 }  } sin\omega t+\frac { 1 }{ \sqrt { 2 }  } cos\omega t)\ \quad =\sqrt { 2 } a(sin\omega t\quad cos\pi /4\quad +cos\omega t\quad sin\pi /4)\ \quad =\sqrt { 2 } a\quad sin(\omega t+\frac { \pi  }{ 4 } )\ This\quad equation\quad satisfy\quad the\quad SHM\quad equation,\quad \frac { { d }^{ 2 }y }{ { d }^{ 2 }t } =-y\quad with\quad amplitude\quad \sqrt { 2 } a\quad .\quad \

Question 2

The equation of motion for an oscillating particle is given by  {x= 3 sin \left ( 4\pi t \right )+4 cos \left ( 4\pi t \right )} , where x is in mm and t is in second

Accepted Answer

(A) x=5\sin (4\pi t+\arctan (4/3))  so motion is simple harmonic (B)  T=2\pi/4\pi=0.5  (C)  x=5\sin (4\pi t+\arctan (4/3))  so amplitude is  5mm (D)  at   t=0, x
eq  0  so false

Question 3

A mass  m= 100\ gms  is attached at the end of a light spring which oscillates on a friction less horizontal table with an amplitude equal to  0.16\ meter  and the time period equal to  2\ sec . Initially the mass is released from rest at  t=0  and displacement  x= -0.16\  meter . The expression for the displacement of the mass at any time  (t)  is

Accepted Answer

Given: m=100 gms\a=0.16m\T=2\ sec Standard equation for given condition  x=a\cos \dfrac{2\pi}{T}t\Rightarrow x=-0.16\cos ( \pi t)

Question 4

Two simple Harmonic Motions of angular frequency 100 and 1000 rad  S^{-1}  have the same displacement amplitude. The ratio of their maximum accelerations is :

Accepted Answer

Maximum acceleration  a =-\omega ^{2}A \dfrac{a_{1}}{a_{2}}=\dfrac{\omega _{1}^{2}A}{\omega _{2}^{2}A}=\dfrac{(100)^{2}}{(1000)^{2}}=\dfrac{1}{10^{2}}

Question 5

The magnitude of maximum acceleration is  \pi   times that of maximum velocity of a simple harmonic oscillator. The time period of the oscillator in seconds is:

Accepted Answer

V_{max} =A \omega a_{max} =\omega ^{2}  A a_{max} =\pi   V_{max} \omega =\pi T=\cfrac{2\pi}{\omega} T=2 \ sec

Question 6

The resultant amplitude due to superposition of two SHMs,   x_{1} =10 sin(\omega t+30^{\circ}  )cm and   x_{2}  =  10 cos (  \omega t+60^{\circ}  )cm is

Accepted Answer

x_{1}=10 sin (wt+30^{0}) x_2=10 cos (wt+60^{0})=-sin(wt-30)=sin(wt+150) Thus, we get the two vectors of magnitude 10 inclined at angles  30^o  and  150^o . The included angle between them is  120^o .Thus the resultant is  \sqrt{10^2+10^2+2(10)(10)cos120}=10cm

Question 7

A  point mass is subjected to two simultaneous sinusoidal displacements in x-direction,  x_{1}(t)=A\sin \omega t  and  x_{2}(t)=A\displaystyle \sin(\omega t+\frac{2\pi}{3})  .  A dding a third sinusoidal displacement  x_{3}(t)=B\sin(\omega t+\phi)  brings the mass to a complete rest. The values of  B  and  \phi  are

Accepted Answer

x_1+x_2 = A\sin \omega t + A\displaystyle \sin(\omega t+\dfrac{2\pi}{3})                   = A\displaystyle \sin(\omega t+\dfrac{\pi}{3}) Since,  x_1+x_2+x_3=0 x_3= A\displaystyle \sin(\omega t+\dfrac{4\pi}{3}) So, B= A  and  \phi =\dfrac{4\pi}{3}

Question 8

Two linear simple harmonic motions of equal amplitude and frequency areimpressed on a particle along x and y-axis respectively. The initial phase difference between them is  \frac{\pi }{2} . The resultant path followed by the particle is :

Accepted Answer

As one is along x- direction and the other along y.Because of initial phase difference  \dfrac{\pi}{2}  it remains same throughout the motion which happens when it travels along a circle.

Question 9

A particle which is simultaneously subjected to two perpendicular simple harmonic motions represented by;  x = \ a_1 cos \omega t  and  y = \ a_2 \ cos 2\omega t  traces a curve given by :

Accepted Answer

Given, y =   {a}_{2}cos \ 2 \omega \  t   =   {a}_{2} (2 \ {cos}^{2} \omega t - 1)  Now, from equation along x axis,   cos \ \omega t = \cfrac{x}{{a}_{1}}  Substituting this value back, y =   {a}_{2}  \ (2 	imes ({\cfrac{x}{ a_1}})^2 - 1)  Hence, this describes a parabola facing upwards.

Question 10

The spring shown in figure is unstretched when a man starts pulling the block. The mass of the block is  m . If the man exerts a constant force  F .The time period of the motion of the block is

Accepted Answer

Here the motion of the block will be simple harmonic. Thus,  F=-kx  where  x  be the displacement of block from equilibrium position. So,  m\dfrac{d^2x}{dt^2}=-kx  or  \dfrac{d^2x}{dt^2}=-(k/m)x Comparing the above equation with general equation of SHM,  \dfrac{d^2x}{dt^2}=-\omega^2x , we get  \omega^2=k/m or  (2\pi /T)^2=k/m So, time period  T=2\pi\sqrt{\dfrac{m}{k}}

Question 11

A mass of  1  kg stretches a spring by  0.8  m. If the mass is pulled down by another  0.05 m and released then the period of oscillation is

Accepted Answer

kx=mg 10=0.8	imes k k=\dfrac{100}{8} T=2\pi \sqrt{8/100}     T=1.77 \ s

Question 12

The mass M shown in the figure oscillates in simple harmonic motion with amplitude A. The amplitude of the point P is

Accepted Answer

Total displacement = Amplitude \mathrm{x}_{1}+\mathrm{x}_{2}=\mathrm{A} Force balance at P \mathrm{k}_{1}\mathrm{x}_{1}=\mathrm{k}_{2}\mathrm{x}_{2} Hence  \displaystyle \mathrm{x}_{1}=\frac{\mathrm{k}_{2}\mathrm{A}}{\mathrm{k}_{1}+\mathrm{k}_{2}}

Question 13

A non viscous liquid of density  \rho  is filled in a tube with A as the area of cross section, as shown in the figure. If the liquid is slightly depressed in one of the arms, the liquid column oscillates with a frequency

Accepted Answer

Liquid of density  =ho Area =A Let, the level of liquid displaced by  x   in one column and in other falls by  x  in another column, then the difference in levels of liquid.Vertical displacement: h=x(\sin	heta_1+\sin	heta_2) 	herefore  Pressure differences P=hho g=ho g(\sin	heta_1+\sin	heta_2)x Now using hook's law in SHM F=-kx , Frequency,        =\cfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\ \Rightarrow f=\cfrac{1}{2\pi}\sqrt{\cfrac{ho g A(\sin	heta_1+\sin	heta_2)}{m}}

Question 14

A small ball of density  \rho _{0}  is released from rest from the surface of a liquid whose density varies with depth  h  as  \rho =\dfrac{\rho _{0}}{2}(a+\beta h) Mass of the ball is  m .Select the most appropriate option:

Accepted Answer

\rho = \dfrac {\rho_{o}}{2} (a + \beta h) V = \dfrac{m}{\rho_{o}} Downward force,  F = mg - \dfrac{m}{\rho _{o}}\dfrac{\rho _{o}}{2}(a+\beta h)g F = mg\left ( \left ( 1- \dfrac{a}{2} \right )- \dfrac{\beta h}{2}\right ) So F is proportional to displacement from center position which is  mg (1 - \dfrac{a}{2})  below surfaceTherefore, the motion is SHM.A =  mg (1 - \dfrac{a}{2}) k = \dfrac{mg \beta}{2} Also,  k = m\omega^{2} =>  \omega = \sqrt{\dfrac{g\beta}{2}} v_{max} = \omega A = \dfrac{(2-a) \sqrt{g}}{\sqrt{2} \beta} Answer A

Question 15

A vibrating system is analysed and it is found that two successive oscillations have amplitudes of  3  mm and  0.5  mm respectively. Calculate the damping ratio.

Accepted Answer

For successive amplitudes  m = 1 Amplitude reduction factor  =ln\left( \dfrac { { x }_{ 1 } }{ { x }_{ 2 } }  ight) =ln\left( \dfrac { 3 }{ 0.5 }  ight) =ln6=1.792 Amplitude reduction factor  =\dfrac { 2\pi \delta m }{ \sqrt { 1-{ \delta  }^{ 2 } }  }  \Rightarrow \dfrac { 2\pi \delta m }{ \sqrt { 1-{ \delta  }^{ 2 } }  } =1.792 squaring both sides \dfrac { 39.478{ \delta  }^{ 2 } }{ 1-{ \delta  }^{ 2 } } =3.21\ \Rightarrow { \delta  }^{ 2 }=0.075\ \Rightarrow \delta =0.274

Question 16

Equation  F=-bv -kx  represents equation of a damped oscillation for a particle of 2 kg mass where  b=\ln2\dfrac{N.S.}{m}  and  k=100  N/m  then time after which energy of oscillations will be reduced to half of initial is :

Accepted Answer

For damped oscillation for a particle of  m=2 \ &#13199; . F=-bV-kx 	herefore   Amplitude ca be written as    A={ A }_{ 0 }{ e }^{ -\cfrac { bt }{ 2m }  } Energy    E\propto { A }^{ 2 } 	herefore   Energy   E={ E }_{ 0 }{ e }^{ -\cfrac { bt }{ m }  } Let after time  { t }_{ 0 }  its energy becomes half of initial energy. 	herefore \cfrac { { E }_{ 0 } }{ 2 } ={ E }_{ 0 }{ e }^{ -\cfrac { b{ t }_{ 0 } }{ m }  } 	herefore ln2=\cfrac { b{ t }_{ 0 } }{ m }  	herefore { t }_{ 0 }=\cfrac { m	imes ln2 }{ b }  	herefore { t }_{ 0 }=\cfrac { 2	imes ln2 }{ ln2 } =2sec 	herefore { t }_{ 0 }=2sec

Question 17

A forced oscillator is acted upon by a force  F={ F }_{ \circ  }sin\omega t. . The amplitude of oscillation is given by  \displaystyle\frac{55}{\sqrt{2\omega^2 - 36\omega + 9}} . The resonant angular frequency is

Accepted Answer

At resonance, amplitude of oscillation is maximum \quad \Rightarrow 2\omega^2 - 36\omega + 9  is maximum \quad \Rightarrow 4\omega - 36 = 0  (derivative is zero) \quad \Rightarrow \omega = 9

Question 18

Amplitude of a wave is represented by   A=\frac{c}{a+b-c}  Then resonance will occur when

Accepted Answer

A=\dfrac{c}{a+b-c}  : When  b = 0, a = cAmplitude  A\rightarrow \infty . This corresponds to resonance

Oscillations